Question 5 - A-Level Maths

CAIE Further Paper 2 2024 November — Question 5 10 marks

Exam Board	CAIE
Module	Further Paper 2 (Further Paper 2)
Year	2024
Session	November
Marks	10
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Second order differential equations
Type	Particular solution with initial conditions
Difficulty	Standard +0.8 This is a standard second-order linear ODE with constant coefficients requiring complementary function (solving auxiliary equation with distinct real roots), particular integral (polynomial trial solution with undetermined coefficients), and applying two initial conditions. While methodical, it's a multi-step Further Maths question requiring careful algebraic manipulation and is more demanding than typical A-level single maths content.
Spec	4.10d Second order homogeneous: auxiliary equation method 4.10e Second order non-homogeneous: complementary + particular integral

5 Find the particular solution of the differential equation $$6 \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } - 5 \frac { \mathrm {~d} x } { \mathrm {~d} t } + x = t ^ { 2 } + t + 1$$ given that, when $t = 0 , x = 12$ and $\frac { \mathrm { d } x } { \mathrm {~d} t } = - 6$.
[0pt] [10] \includegraphics[max width=\textwidth, alt={}, center]{4af32247-c1f9-4c1f-bdf8-bafe17aca1dc-10_2715_40_110_2007} \includegraphics[max width=\textwidth, alt={}, center]{4af32247-c1f9-4c1f-bdf8-bafe17aca1dc-11_2726_35_97_20}

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Question 5:

Answer	Marks	Guidance
$6m^2-5m+1=0 \Rightarrow m=\frac{1}{2},\frac{1}{3}$	M1	Auxiliary equation, two distinct real roots.
$x = Ae^{\frac{1}{2}t}+Be^{\frac{1}{3}t}$	A1	Complementary function.
$x = pt^2+qt+r \Rightarrow x'=2pt+q \Rightarrow x''=2p$	B1	Particular integral and its derivatives.
$p=1 \quad -10p+q=1 \quad 12p-5q+r=1$	M1	Substitutes and equates coefficients.
$p=1 \quad q=11 \quad r=44$	A1
$x = Ae^{\frac{1}{2}t}+Be^{\frac{1}{3}t}+t^2+11t+44$	A1	General solution. Need to see '$x=$'. FT on CF.
$x' = \frac{1}{2}Ae^{\frac{1}{2}t}+\frac{1}{3}Be^{\frac{1}{3}t}+2t+11$	M1	Differentiates.
$A+B+44=12 \quad \frac{1}{2}A+\frac{1}{3}B+11=-6 \Rightarrow A=-38,\ B=6$	M1A1	Uses initial conditions.
$x = -38e^{\frac{1}{2}t}+6e^{\frac{1}{3}t}+t^2+11t+44$	A1	Need to see '$x=$'.

Total: 10

## Question 5:

$6m^2-5m+1=0 \Rightarrow m=\frac{1}{2},\frac{1}{3}$ | M1 | Auxiliary equation, two distinct real roots.

$x = Ae^{\frac{1}{2}t}+Be^{\frac{1}{3}t}$ | A1 | Complementary function.

$x = pt^2+qt+r \Rightarrow x'=2pt+q \Rightarrow x''=2p$ | B1 | Particular integral and its derivatives.

$p=1 \quad -10p+q=1 \quad 12p-5q+r=1$ | M1 | Substitutes and equates coefficients.

$p=1 \quad q=11 \quad r=44$ | A1 |

$x = Ae^{\frac{1}{2}t}+Be^{\frac{1}{3}t}+t^2+11t+44$ | A1 | General solution. Need to see '$x=$'. FT on CF.

$x' = \frac{1}{2}Ae^{\frac{1}{2}t}+\frac{1}{3}Be^{\frac{1}{3}t}+2t+11$ | M1 | Differentiates.

$A+B+44=12 \quad \frac{1}{2}A+\frac{1}{3}B+11=-6 \Rightarrow A=-38,\ B=6$ | M1A1 | Uses initial conditions.

$x = -38e^{\frac{1}{2}t}+6e^{\frac{1}{3}t}+t^2+11t+44$ | A1 | Need to see '$x=$'.

**Total: 10**

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5 Find the particular solution of the differential equation

$$6 \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } - 5 \frac { \mathrm {~d} x } { \mathrm {~d} t } + x = t ^ { 2 } + t + 1$$

given that, when $t = 0 , x = 12$ and $\frac { \mathrm { d } x } { \mathrm {~d} t } = - 6$.\\[0pt]
[10]\\

\includegraphics[max width=\textwidth, alt={}, center]{4af32247-c1f9-4c1f-bdf8-bafe17aca1dc-10_2715_40_110_2007}\\
\includegraphics[max width=\textwidth, alt={}, center]{4af32247-c1f9-4c1f-bdf8-bafe17aca1dc-11_2726_35_97_20}\\

\hfill \mbox{\textit{CAIE Further Paper 2 2024 Q5 [10]}}

This paper (7 questions)

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Q2 6 Q3 12 Q4 9 Q5 10 Q6 10 Q7 10 Q8 14

Answer	Marks	Guidance
\(6m^2-5m+1=0 \Rightarrow m=\frac{1}{2},\frac{1}{3}\)	M1	Auxiliary equation, two distinct real roots.
\(x = Ae^{\frac{1}{2}t}+Be^{\frac{1}{3}t}\)	A1	Complementary function.
\(x = pt^2+qt+r \Rightarrow x'=2pt+q \Rightarrow x''=2p\)	B1	Particular integral and its derivatives.
\(p=1 \quad -10p+q=1 \quad 12p-5q+r=1\)	M1	Substitutes and equates coefficients.
\(p=1 \quad q=11 \quad r=44\)	A1
\(x = Ae^{\frac{1}{2}t}+Be^{\frac{1}{3}t}+t^2+11t+44\)	A1	General solution. Need to see '\(x=\)'. FT on CF.
\(x' = \frac{1}{2}Ae^{\frac{1}{2}t}+\frac{1}{3}Be^{\frac{1}{3}t}+2t+11\)	M1	Differentiates.
\(A+B+44=12 \quad \frac{1}{2}A+\frac{1}{3}B+11=-6 \Rightarrow A=-38,\ B=6\)	M1A1	Uses initial conditions.
\(x = -38e^{\frac{1}{2}t}+6e^{\frac{1}{3}t}+t^2+11t+44\)	A1	Need to see '\(x=\)'.